solar lead acid battery controller

walle62 · 2005-06-09T11:22:24+00:00

dear friendsi design a solar lead acid battery controller.it feautures:battery nominal voltage: 17 V surge load current: 15A or 30Ait does not need a regulatorit should protect against:over-, undercharge, short circuit, solar cells reverse current, batteryand load reverse polarityit must contain a gas gauge (battery charge measurement)i must have a self explaining display of the battery charge in Ah andof the charge current in Athe design must be simple and cost effectivemy idea is to use bipolar transistors for measurement of the over-,undervoltage etc. and drive power-mos-fets. or exists a good ic thatfeatures all conditions? perhaps i could also use a multi-op-amp deviceas the LM 324 and c-mos-digital-ics for threshold, driving etc.but i have no idea how to implement the gas gauge. can i determine thebattery charge via battery voltage?i would be very glad if one of you could help or give me a hint!thank you very much!if i find a good design, i will report you here in the forum!with kind regardswalter nussbaum, zuerich, switzerland

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mike 21 years ago

Sounds like you have more experience than I do. I understnd about the short-circuit current. Does the same apply to current into a 14V battery? I did some experiments with a small solar cell. I determined that the available charge current went down dramatically for small changes in incident angle. If you don't track the sun, there's a significant part of the day when the EFECTIVE light input is way down.

Yep, all you need is a transistor and a big resistor to ground. Simple. But when you do the math, 15V max voltage 40A max shunt current, the maximum power in your transistor is 7.5V X 20A = 150W.

What we actually did for the mountain top repeater was to build a

10A switching shunt and switched additional 10A resistors in as needed. Once we had the pic for voltage measurement, all the other stuff became trivial. Very low power in the switching FETS and BIG resistors mounted on the wall to dissipate the heat. Also limits the surge currents in your batteries without having to add inductors or caps. The pic can also let you correct voltage points for temperature. BIG temp changes on a mountain top. Been running about a year now without a hitch.

Oh, there was another complication. This site also had wind power. Couldn't just disconnect the load with the wind at 60MPH.

I'll need more explanation of this. I always considered the inductor loss to be mostly due to it's resistance and therefore the current. For the diode, a diode drop is a diode drop. Yes, there are also dynamic losses. What's the voltage swing at the input to the inductor when the input and output voltages are close?

mike

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Ken Smith 21 years ago

[...]

I'm not sure I understand the question but I'll take a shot at it.

A fully discharged 12V lead acid battery is right about 10.5V. A fully charged one is about 13.7V.

To be able to fully charge the battery, the solar cells need to be able to make something like 13.7V.

In a simple system, enough solar cells are connected in series to charge the battery and connected nearly directly to it until the battery is charged. In the short term, the terminal voltage of the lead acid battery is nearly independant of the charging current (ie: it has a low impedance). This will work to your advantage if you are making a switcher for the charging circuit. If you use a PIC, you can dither the PWMing up and down to find the point where the greatest current is flowing into the battery.

Yes it falls of with angle faster than a cos() curve. The sun hitting the face of the panel follows a cos() curve. The percentage that refects back away from the surface increases with increasing angle too.

If you point the panel towards the sun, there will be about 1KW of power falling on a square meter of panel. 10% of the power can end up as electrical power. The rest ends up as heat.

It comes down to a question of cost whether tracking makes sense or not. Any money you spend on gears and motors could also be spent on another panel. You need to price the situation both ways to see which is better. At one time the motors and gears where always the lower cost way to go. I don't know about today.

[...]

I'd be inclined to use many TO-220 resistors for the power resistor. At

20W each, they would spread the heat around nicely. [....]

The "real world" is never as nice as the lab bench.

[...]

The AC resistance of an inductor is always more than the DC resistance and the more the difference between the input and output, the more inductance you need.

This means that the losses in the design, end up determined by the input to output difference. (If the difference was zero, the inductance would be zero and so could the resistance)

Well not quite. If you don't use a very fast diode, the storage effect adds to the losses. Also remember that the diode only conducts for a small fraction of the total cycle.

Until they are almost exactly equal, the peak swing remains the input voltage. The RMS of the AC part of the waveform decreases as you near equal. There is a bit of a bummer in that the high harmonics decrease slower. The losses at the higher harmonics are proportionally larger.

-- kensmith@rahul.net forging knowledge

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Ross Herbert 21 years ago

Yes. In large battery installations such as in telephone exchanges the most common cell used was a 2V, 1000 or 2000Ah flooded lead acid type configured as 24 cells in series. Despite the fact that all cells were from the same manufacturing batch when installed it was not uncommon for a single cell to lose capacity after a few years of service which then had to be replaced. The only way to detect the weak cell was to perform periodic testing of individual cells.

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